been trying to do an efficient vlookup style on pandas, with IF function...

Basically, I want to apply to this column ccy_grp, that if the value (in a particular row) is 'NaN', it will take the value from another column ccy

def func1(tkn1, tkn2):
    if tkn1 == 'NaN:
        return tkn2
    else:
        return tkn1
    
tmp1_.ccy_grp = tmp1_.apply(lambda x: func1(x.ccy_grp, x.ccy), axis = 1)

but nope, doesn't work. The code cannot seem to detect 'NaN'. I tried another way of np.isnan(tkn1), but I just get a boolean error message...

Any experienced python pandas code developer know?

CodePudding user response：

use pandas.isna to detect a value whether a NaN

1. generate data

import pandas as pd
import numpy as np

data = pd.DataFrame({'value':[np.NAN, None, 1,2,3],
                     'label':['str:np.NAN', 'str: None', 'str: 1', 'str: 2', 'str: 3']})
data

2. create a function

def func1(x):
    if pd.isna(x):
        return 'is a na'
    else:
        return f'{x}'

3. apply function to data

data['func1_result'] = data['value'].apply((lambda x: func1(x)))
data

CodePudding user response：

There is a pandas method for what you are trying to do. Check out combine_first:

Update null elements with value in the same location in ‘other’.

Combine two Series objects by filling null values in one Series with non-null values from the other Series.

tmp1_.ccy_grp = tmp1_.ccy_grp.combine_first(tmp1_.ccy)

CodePudding user response：

This looks like it should be a pandas mask/where/fillna problem, not an apply:

Given:

   value  values2
0    NaN      0.0
1    NaN      0.0
2    1.0      1.0
3    2.0      2.0
4    3.0      3.0

Doing:

df.value.fillna(df.values2, inplace=True)
print(df)

# or

df.value.mask(df.value.isna(), df.values2, inplace=True)
print(df)

# or 

df.value.where(df.value.notna(), df.values2, inplace=True)
print(df)

Output:

   value  values2
0    0.0      0.0
1    0.0      0.0
2    1.0      1.0
3    2.0      2.0
4    3.0      3.0